You need to express the coordinate basis vectors for the current system in terms of a linear combination of the coordinate basis vectors for spherical coordinates, and substitute into your equation. Alternatively, there is a formula for directly converting the components of your tensor from one coordinate system to another. Both methods give you the same answer.
Method 1: $$\vec{i}_x=\sin{\theta}\cos{\phi}\vec{i}_r+\cos{\theta}\cos{\phi}\vec{i}_{\theta}-sin{\phi}\vec{i}_{\phi}$$ $$\vec{i}_y=\sin{\theta}\sin{\phi}\vec{i}_r+\cos{\theta}\sin{\phi}\vec{i}_{\theta}+cos{\phi}\vec{i}_{\phi}$$ $$\vec{i}_z=\cos{\theta}\vec{i}_r-\sin{\theta}\vec{i}_{\theta}$$
I'm only going to do it for a simple case in which only one component of the stress tensor is non-zero in cartesian coordinates, the z-z component. So,
$$\vec{\sigma}=\sigma_{zz}\vec{i}_z \otimes \vec{i}_z=\sigma_{zz}(\cos{\theta}\vec{i}_r-\sin{\theta}\vec{i}_{\theta})\otimes(\cos{\theta}\vec{i}_r-\sin{\theta}\vec{i}_{\theta})$$ So, $$\vec{\sigma}=\sigma_{zz}\vec{i}_z \otimes \vec{i}_z=\sigma_{zz}(cos^2\theta\vec{i}_r \otimes \vec{i}_r-sin\theta cos\theta(\vec{i}_r \otimes \vec{i}_{\theta}+\vec{i}_{\theta} \otimes \vec{i}_r)+sin^2\theta\vec{i}_{\theta} \otimes \vec{i}_{\theta})$$ So, in this case, it follows that: $$\sigma_{rr}=\sigma_{zz}cos^2\theta$$ $$\sigma_{r\theta}=\sigma_{\theta r}=-\sigma_{zz}sin\theta cos\theta$$ $$\sigma_{\theta \theta}=\sigma_{zz}sin^2\theta$$